Mathematics in the first year of high school is about recursive sequences, helping to pull the urgen

1)3a(n+1)^2-3=2an^2-2a(n+1)^2-1=2/3(an^2-1)a1=1???

an=1 if a1=k is not equal to 1

Let bn=an 2-1

then b1=k 2-1

b(n+1)=2/3bn

bn=b1(2/3)^(n-1)=(k^2-1)*(2/3)^(n-1)

an=√[(k^2-1)*(2/3)^(n-1)+1]2)1/√n+√(n+1)= √(n+1)-√na(n+1)-√n+1)=an-√n

an=√n+k

Because a1 = 1

So a1=1+k=1, k=0

So an= n

3) If it is na(n+1)=(n+1)an

then a(n+1) (n+1)=an n=k

Because a1 = 1

So k=1, so an=n

4)a(n+1)+1=-2an-2=-2(an+1)a(n+1)+1]/(an+1)=-2

an+1=k+(-2)^n

a1=12=k-2

k=4an=3+(-2)^n

5) 3a(n+1) 2+2ana(n+1)-an 2=03a(n+1)-an][a(n+1)+an]=0 can be 3a(n+1)=an or a(n+1)=-(an)6)sn=2an-2

s(n-1)=sn-an=an-2=2a(n-1)-2, so an=2a(n-1).

There are so many questions, I can only tell you the way of thinking.

1.Subtract 3 on both sides

Move to the left, and then use the accumulation method, and the denominator is rationalized.

3.a(n+1) an=n (n+1) and multiplication.

4.And the first question is of the same type, with 1 and 3 on both sides

5.Why is there no equal sign in this question?

s(n-1)=2a(n-1)-2 subtract the two, but don't forget the case when n=1.

These questions are the most basic number series questions, and there will be more difficult ones in the future, if there are any number series questions that you can't ask, you can ask me, I have learned the number series well.

Ever heard of characteristic equations? Many series of recursive can be used to construct equal difference or proportional series, which is also the basic way to find the general term of the series recursively!

I'll solve this problem first, the key is to subtract 1 from the left and right, and then reverse it. This minus one is also traceable - the characteristic equation!

Solution: (Note: I use a [n] to avoid confusion).

by a[n+1]=1 (2-a[n]).

A[n+1]-1=1 (2-a[n])-1

i.e. a[n+1]-1=(a[n]-1) (2-a[n]).

So 1 (a[n+1]-1)=(2-a[n]) a[n]-1).

i.e. 1 (a[n+1]-1)=-1+1 (a[n]-1).

So it is a series of equal differences with 1 (a-1) as the first term and -1 as the tolerance.

So 1 (a[n]-1)=1 (a-1)-(n-1).

So a[n] = [a-(n-1)(a-1)] 1-(n-1)(a-1)]].

It seems to be very magical, this step-1 is actually obtained from the eigenroot: replace a[n] and a[n+1] with x to get the eigenequation: x=1 (2-x) a quadratic equation, get the solution (such as this x=1) left and right sides subtract the obtained solution, reciprocal, and then it can be turned into a recursive formula similar to the above about a[n] withering coefficient, which may be equal difference, or it can be equal to the failure of the panicle, after the construction, find the general term of the number series created by the solution, and then turn out a[n].

The general characteristic equation is a solution to produce an equal difference, and the two solutions are equal proportions (two solutions are optional), but it is also possible to choose two formulas and divide them to obtain an equal proportional series. The above is the reciprocal solution. It is called the characteristic equation or the fixed point method.

Or I'll give you a practice: a[n+1]=2 (a[n]-1)

Add up the equation of all the empty and the empty and the land.

To the left of the equation is.

a2+..a(n-1)+a(n)

On the right. a1+a2+..a(n-1)+f(1)+.f(n-1) eliminates a2+...a(n-1).

a(n)=a1+f(1)+.f(n-1)

The first question is all this kind of derivation.

can be disassembled.

an+x=(a(n+1))before coefficients) times (a(n+1)+x), at this time it becomes an equal proportional series, (a (n+1)+x) (an+x) = (a(n+1) before coefficients) such a formula to push, the premise, the coefficient before an must be 1, if it is not one, it must become one first, and then set the formula, you try it yourself, you will have a very deep understanding, and you will solve it like this in the future.

Second question. I'm sorry.

Can't see what you're writing clearly.

The third question, the law is found, which requires a general formula.

It can be done 100% by mathematical induction

In general, to prove a proposition p(n) related to the natural number n, there are the following steps:

1) Prove that the proposition is true when n takes the first value n0. n0 is 0 or 1 for a general series, but there are special cases;

2) Suppose when n=k(

k n0, k is a natural number.

The proposition is true, proving that the proposition is also true when n=k+1.

Synthesis (1) (2), for all natural numbers n( n0), the proposition p(n) holds. Hope. Thank you.

There is nothing I understand.

**Friends. Answer them one by one.

Add all the equations together.

To the left of the equation is.

a2+..a(n-1)+a(n)

On the right. a1+a2+..a(n-1)+f(1)+.f(n-1) eliminates a2+...a(n-1).

a(n)=a1+f(1)+.f(n-1)

Formula method, accumulation method, accumulation method, undetermined coefficient method, logarithmic transformation method, iterative method, mathematical induction method, commutation method, fixed point method, eigenroot method, etc.

Type 1: Induction-Conjecture-Proof.

From the recursive formula of the number series, the first few terms of the number series can be written, and then the law can be summarized from the first few terms, and a general term formula of the number series can be guessed, and finally proved by mathematical induction

Type 2: "Difference-by-Difference" and "Accumulation Method".

1) When the recursive formula of the series can be reduced to an+1-an=f(n), take n=1,2,3 ,..., n-1, get n-1 formula:

a2-a1=f(1),a3-a2=f(2),…an-an-1=f(n-1), and f(1)+f(2)+....When f(n-1) can be obtained, the two sides are added up to obtain the general term an, which is called the "difference method".

2) When the recursive formula of the series can be changed to an+1 an=f(n), let n=1,2,3 ,..., n-1, get n-1 formula, i.e.

a2/a1=f(1),a3/a2=f(2),a4/a3=f(3),…an an-1=f(n 1), and f(1)f(2)f(3)....When f(n-1) can be obtained, an can be obtained by multiplying the two sides, and this method is called the "product business method".

Type 3 construction method.

The recursive formula is pan=qan-1+f(n) (p and q are non-zero constants), and a new proportional series can be constructed by the method of undetermined coefficients

Type 4 can be converted to type 3 to find the general term.

1) "Logarithmic method" is transformed into type three

The recursive formula is an+1=qan

This needs to be summarized by yourself in the question.

If I told you all the methods.

Will you understand?

You do one question at a time.

Just summarize what type of series this belongs to.

Hope it helps.

a1=2a2=a1+3*1+2=7

a3=a2+3*2+2=15

Sequence 2, 7, 15, 1, analytical.

Suppose an+1=an+2

a1=2a2=2+2

a3=2+2+2

There can be an=2n

Let's assume that an+1=an+3n

a1=2a2=2+1*3

a3=2+1*3+2*3

a4=2+3*1+2*3+3*3

an=2+3(1+2+3+4+..n-1)=2+3(n-1)n2 is assumed twice.

then an+1=an+3n+2

an=2n+3(n-1)n/2=（n+n^2) /2 =(3n+1)n/2

2. Iterative method.

an+1=an +3n+2

an=an-1 +3(n-1)+2

an-2 +3(n-1) +3(n-2)+2an-3 +3(n-1) +3(n-2)+2 +3(n-3)+2a1 +3(n-1)+2 +3(n-2)+2 +.3(n-(n-1))+2

2n+ 3n(n-1)-3(1+2+3+..n-1)4n/2+ (6n^2-6n)/2- 3n(n-1)/2(3n+1)n/2

3. Dislocation method.

Make a list of the first few items in the series.

It can be seen that it is a second-order difference series.

a2-a1=5 =3*1+2

a3-a2=8=3*2+2

a4-a3=11=3*3+2

an- an-1=3*(n-1) +2

All of the above equations are added.

an -a1=3(2+3+4+..n-1)+2(n-1)an=(3n+1)n/2

4. Substitution method.

an=2c(n,0)+5c(n,1)+3c(n,2) I hate myself.

a1=2a2-a1=3*1+2

a3-a2=8=3*2+2

a4-a3=11=3*3+2

an-a(n-1)=3*(n-1)+2

There is addition. an=3*(1+2+..n-1)+2*n3*n(n-1)/2+2n

3n^2-3n+4n)/2

3n^2+n)/2

Substituting n=1, n=2 checks. The answer is correct.

Solution: (1).

s1=a1=2a1-3

a1=3sn=2an-3n

sn-1=2a(n-1)-3(n-1)

an=sn-sn-1=2an-3n-2a(n-1)+3(n-1)=2an-2a(n-1)-3

an=2a(n-1)+3

an+3=2a(n-1)+6=2[a(n-1)+3](an+3) [a(n-1)+3]=2, which is a fixed value.

a1+3=3+3=6

The number series is a proportional series with 6 as the first term and 2 as the common ratio.

an+3=6×2^(n-1)=3×2^n

an=3(2 n-1) (parentheses are 2 to the nth power minus 1).

When n=1, a1=3 (2-1)=3, is equally satisfied.

The general formula for the series is an=3(2 n-1).

2) Suppose that there are three items in the series that satisfy the topic, let these three items be ap,aq,at2aq=ap+at

6(2^q-1)=3(2^p+2^t-2)2^(q+1)=2^p+2^t

2 (p-q-1)+2 (t-q-1)=1 The left side of the equation is an even number, and the right side is an odd number, and the equation does not hold.

That is, there are no three terms in the sequence that can form a series of equal differences.